PreCal please help!
A population of bacteria is growing according to the equation . Estimate when the population will exceed 493.

P(t)=400e^(0.05t)

Question

PreCal please help! 
A population of bacteria is growing according to the equation . Estimate when the population will exceed 493.

P(t)=400e^(0.05t)

anonymous · Answer

So I know my equation is 493=400e^(0.05t) but I can't figure out how to solve it

anonymous · Answer

493 = 400*e^(0.05t)
$$\ln (493/400) = 0.05t$$$$t=20*\ln (493/400) = 4.18$$

anonymous · Answer

where did you get 20 from? we haven't learned about natural log yet. I think that's what ln is...

anonymous · Answer

You are correct, ln is natural log :D  It's just a special form of a logarithm with base "e"
1/.05 is 20 :D